42 Summer 2015 The BenT of Tau BeTa pi

Brain Ticklers

RESULTS FROMWINTER

WINTER REVIEW

Winter #4 about the maximum num-ber of cards in a 5-Spot deck was as difficult as the Bonus and kept seven entries out of the "Perfect" section.

SPRING SOLUTIONS

Readers’ entries for the Spring Ticklers will be acknowledged in the Fall Bent. Meanwhile, here are the answers.

Perfect *Bohdan, Timothy E. IN Γ ’85*Gerken, Gary M. CA H ’11* Jones, John F. WI Α ’59 Jones, Jeffrey C. Son Richards, John R. NJ B ’76 Schmidt, V. Hugo WA B ’51*Spong, Robert N. UT Α ’58*Strong, Michael D. PA Α ’84*Voellinger, Edward J. Non-member

Other Alexander, Jay A. IL Γ ’86*Brule, John D. MI B ’49*Gaston, Charles A. PA B ’61*Gibbs, Kenneth P. MO Γ ’76 Grewal, Rashi NJ Γ ’09*Gulian, Franklin J. DE Α ’83 Gulian, William Son Handley, Vernon K. GA Α ’86 Jones, Donlan F. CA Ζ ’52 Kiefer, Thomas G. CA B ’91 Kimsey, David B. AL Α ’71 Lalinsky, Mark A. MI Γ ’77 Marks, Lawrence B. NY Ι ’81 Hertz, Caryn M. NY Ι ’81 Marks, Benjamin Son Marrone, James D. IN Α ’87 Marrone, Julia Daughter Marrone, Louisa Daughter Marrone, James I. IN Α ’61 Norris, Thomas G. OK Α ’56 Pendleton III, Winston MI Γ ’62 Quan, Richard CA Χ ’01 Rentz, Peter E. IN Α ’55 Riedesel, Jeremy M. OH B ’96 Saikali, Jeffrey G. OH B ’96 Shapiro, Scott E. IL Α ’12 Kovarik, Ryan Non-member Sigillito, Vincent G. MD B ’58 Siskind, Kenneth S. RI Α ’86 Siskind, Amanda M. Daughter Siskind, Brian A. Son Soni, Yash P. MA H ’16*Stribling, Jeffrey R. CA Α ’92 Summerfield, Steven L. MO Γ ’85 Vinoski, Stephen B. TN D ’85 Yacobucci, Roger B. CO D ’88 Yacobucci, Daniel CO D ‘10 Yacobucci, Nancy Faye Daughter

*Denotes correct bonus solution

B = x2 and (E2) 100(A+13) + B + 13 = y2, or on expanding, 100A+B+1313 = y2. Subtracting (E1) from (E2) gives y2 – x2 = (y–x)(y+x) = 1313 = 13(101). Therefore, y–x=13 and y+x=101. Solving gives x=44 and y=57, with 442=1936 and 572=3249. So, Alice is 19 (32 in 13 years) and Bill is 36 (49 in 13 years).

3 It takes at least 11 crossings to get the three couples safely across the river. The three right hand col-umns in the table list the individuals on the near bank, in the boat, and on the far bank for each crossing.

4 There is a 0.25 probability that your friend (call him Bob) took a blue marble. There are two possi-bilities: Bob took red and you took blue, or Bob and you both took blue. Initially, the bag has 3R and 2B, so there is a 0.6 probability that Bob chose red and a 0.4 probability that he chose blue. You might assume, therefore, that Bob’s probability of getting a B is 0.4 and that is the answer, but the fact that you drew a B decreases the probability that Bob got a B. If he chose red, that would leave 2R and 2B, and you would have a probability of 0.5 of picking B. Thus, the probability of Bob R, you B is 0.5(0.6) = 0.3. If Bob picked B, then 3R and 1B would be left, and your probability of drawing B would be only 0.25, so the probability of Bob B, you B is 0.25(0.4) = 0.1, and the relative probability is 0.1/(0.1+0.3) = 0.25.

1 DONATE – MONEY – TO – THE + NEW = ENDOW + MENT translates to 129034-72945-32-384+946=49126+7493. Moving the negative terms to one side and the positive terms to the other and then canceling letters which occur on both sides in the same columns (for example, the W in NEW and EN-DOW) and consolidating gives:

E M D Ø T E N Y M Ø T H ØD Ø 0 A 0 0

Summing the individual columns produces the following five equa-tions:

T +Y+Ø=10Ø+N+H+1=20D+E+T+2=A+10M+Ø+1=10E+M+1=Ø+10

First, we see that D=1. Next, Ø=9-M, and E+M=Ø+9, or E+M=9-M+9, so E+2M=18, so possible values for (E, M, Ø) are (8, 5, 4), (4, 7, 2) and (2, 8, 1), but this last possibility must be rejected because D=1, so Ø=2 or 4. Now, N+H=19-Ø, and T+Y=10-Ø. If we assume Ø=4, then T+Y=10-Ø=6, so (T, Y)=(1, 5) or (2, 4) in some order, but D=1 and Ø=4, so Ø cannot equal 4 and must be 2. Then, we have: D=1, Ø=2, M=9- Ø=7, E=18-2M=4, T+Y=10-Ø=8, so (T, Y)=(3, 5) or (5, 3); they cannot be (1, 7) or (2, 6) as 1 and 2 are already assigned; since N+H=19- Ø=17, so (N, H) must be (8, 9) in some order, A=D+E+T-10=1+4+T-10. Since A cannot be negative, T=5, A=0, and Y=3. Finally, W must equal 6. This gives (A, D, E, H, M, N, Ø, T, W, Y)=(0, 1, 4, 8/9, 7, 9/8, 2, 5, 6, 3). Of the two possibilities, N=9 gives the larger ENDOWMENT== 491267493 and produces the answer given above.

2 Ann and Bill’s ages are 19 and 36 (or vice versa). Let A = Ann’s age and B = Bill’s age. Then, (E1) 100A +

43Summer 2015 The BenT of Tau BeTa pi

N(2N–1)/(6N2) = N + k(N–1)/2 – k(N–1)(2N–1)/(6N) = N + k[(N–1)/2][1 – (2N–1)/3N] = N + k[(N–1)/2][(N+1)/3N] = N + k(N2 – 1)/(6N).

Bonus 167 is the smallest magic sum for a magic square consisting entirely of semiprimes and meeting the specified conditions. One such square is:

09 35 85 38

49 74 10 34

15 25 58 69

94 33 14 26

A novel approach is to find a 4×4 ar-ray of 16 semiprimes with the prop-erty that the differences between entries in the 1st and 2nd columns, between the 1st and 3rd columns, and between the 1st and 4th columns are constants (call them D1, D2, and D3, respectively), with a similar situa-tion for the rows (call the row con-stants δ1, δ2, and δ3, respectively). The following array, found by in-spection of a listing of all semiprimes less than 100, has this property, with D1=5, D2=16, D3=25, δ1=1, δ2=24, and δ3=60:

(1) (2) (3) (4)A: 9 14 25 34B: 10 15 26 35C: 33 38 49 58D: 69 74 85 94 Now, consider a 4×4 Greco-Roman square, such as:

A1 B2 C3 D4

D3 C4 B1 A2

B4 A3 D2 C1

C2 D1 A4 B3

where each letter and each number appear once in each row, column, and major diagonal, and no letter/number pair is duplicated. There are two fundamentally different 4×4 Greco-Roman squares. If we insert the corresponding members from the semiprime array into the Greco-Roman square, i.e., A(1) into

A1, A(2) into A2, etc., we obtain the desired magic square shown above. A property of the array of semiprimes above is that each en-try can be expressed as a function of the first entry and the deltas, as shown below. If we let n=9 and the deltas have the values listed above, we get the magic square above. If you substitute from the following table, you get a magic sum of S = 4n+D1+D2+D3+δ1+δ2+δ3 = 4(9)+5+16+25+1+24+60 = 167.

n n+D1 n+D2 n+D3

n+δ1 n+D1+δ1 n+D2+δ1 n+D3+δ1

n+δ2 n+D1+δ2 n+D2+δ2 n+D3+δ2

n+δ3 n+D1+δ3 n+D2+δ3 n+D3+δ3

We can generate variations of the magic square by permuting the let-ters and numbers of the Greco-Ro-man square. Since there are 4 letters and 4 numbers and 2 squares, there are 2(4!)2 = 1152 possible magic squares, which, upon eliminating ro-tations and reflections, gives 1152/8 = 144 fundamentally different magic squares with a magic sum of 167.

Computer Bonus The smallest social number of degree 28 is 14,316. This Tickler requires a program to get the prime factorization of a num-ber N = p1

ap2bp3

c…pin. Then,

the sum of the divisors is given by the following product, which hasone factor for each different prime: ∏n

k=1[(pkm+1-1)/(pk-1)] - N. The expo-

nent m is that associated with prime pk in the prime factorization. Try successive numbers until you find one that returns to its original sum on the 28th cycle; with 14,316, the next sum is 19,116, and the sums gradually increase to 629,072, then slowly decrease back to 14,316.

1 There was a young and adventur-ous man who found among his great-grandfather's papers a map on a piece of parchment that revealed the location of a hidden treasure. The instructions read:

NEW SUMMER PROBLEMS

5 For a 100 story building in the shape of a square antiprism, the ratio, R, of total floor space to the square base is 113.8. The general equation for an N story building is R = N + k(N2 – 1)/(6N), where k = 2(√2-1). The cross section of the OWTC at any level is an octagon with alternating longer and shorter

sides; only at the exact middle is the cross section a regular octagon. Let L be the length of a side of the

square base at each end of the anti-prism. Define x as the length of each of four sides of an octagonal cross section that starts from L at the base and narrows to 0 at the top. Let y be the length of the alternate four sides of the octagon that vary from 0 at the base to L at the top. Let a and b be the lengths of the perpendiculars from the center of the octagon to sides of lengths x and y respectively. Now, x = (1 – f)L, where f is the fraction of the building’s height at which the cross section is located, and y = fL. Now, a goes from L/2 at the bottom to L√2/2 at the top, so a = L/2 + (L√2/2 – L/2)f = (L/2)[1 + (√2 – 1)f], and b = (L/2)[1 + (√2 – 1)(1 – f)]. The area of A1 = ax/2 = (L/2)[1 + (√2 – 1)f](1 – f)L/2 = (L2/4)[1 + (√2 – 1)f – f – (√2 – 1)f2], and the area of A2 = by/2 = (L/2)[1 + (√2 – 1)(1 – f)]f/2 = (L2/4)[f + (√2 – 1)f – (√2 – 1)f2]. The area of a cross section is At = 4(A1 + A2) = L2[1 + 2(√2 – 1)f – 2(√2 – 1)f2 = L2(1 + kf – kf2) = L2[1 + k(f – f2)]. Based on this equation and varying f from 0 to 0.99 in increments of 0.01, a spreadsheet can quickly calculate the ratio of the floor space of a 100 story building to the square base as 113.8. To get a general equation, we must sum the floor space from 0 to N – 1, where N is the number of stories. To start, replace f with n/N and, since we want a ratio to the base, we can let L = 1. Therefore, we have S = ∑0

N-1

[1 + k(n/N – n2/N2)], which can be evaluated using the formulas for the summation of integers [n(n+1)/2] and squares [n(n+1)(2n+1)/6] to give R = N + k(N–1)N/(2N) – k(N–1)

44 Summer 2015 The BenT of Tau BeTa pi

"Sail to ___ latitude and ___ longi-tude where thou wilt find a deserted island. [When this Tickler first ap-peared in a 1955 Bent, the island’s location was given as 16oN 16oW, but that is obviously incorrect, and the true coordinates, which were known to the young man at the time he found the map, have since been lost.] There lieth a large meadow, not pent, on the north shore of the island where standeth a lonely oak and a lonely pine. There thou wilt see also an old gallows on which we once were wont to hang traitors. Start thou from the gallows and walk to the oak counting thy steps. At the oak thou must turn right by a right angle and take the same number of steps. Put here a spike in the ground. Now must thou re-turn to the gallows and walk to the pine counting thy steps. At the pine thou must turn left by a right angle and see that thou takest the same number of steps, and put another spike into the ground. Dig halfway between the spikes; the buried trea-sure is there." The instructions were quite clear and explicit, so our young man char-tered a ship and set sail. He found the island, the field, the oak and the pine, but to his great sorrow the gallows were gone. Too long a time had passed since the document had been written; rain and sun and wind had disintegrated the wood and re-turned it to the soil, leaving no trace even of the place where it once had stood. The young man, not having been a Tau Bate, gave up in despair. Perhaps YOU can find the gallows' former location for him.

—One, Two, Three...Infinity by George Gamow

2 Consider five points arranged as the vertices of a regular pentagon and labeled A through E. Identical resistors are used to connect each of the following pairs of points: A-B, A-C, A-D, B-C, B-D, B-E, C-D, C-E, and D-E. Note that the pair A-E is not connected. The resistance between points A and E is 1 ohm. When three of the nine resistors are removed, the resistance between points A and E remains 1 ohm.

Which resistors are removed, and what is the value of one of the resis-tors?

—Hubert W. Hagadorn, PA E ’59

3 A new theater, with exactly 200 seats, has just opened. Every night 200 people, each with an assigned seat, line up. However, the first person in line is an acoustic engi-neer, who is interested in assessing the sound quality throughout the theater. The engineer picks a seat at random, which may or may not be his assigned seat. As the rest of the people enter in order, if their as-signed seat is vacant, they sit in it. If not, they pick any unoccupied seat. What is the exact probability that the last person in line sits in their as-signed seat?

—The Everything Brain Strain Book by Jake Olefsky

4 The DryFroot Company makes one pound (16 oz) boxes of Froot-Mix. Each box is made to the same recipe and has a positive integral number of ounces of apples, bananas, cranberries and dates. The com-pany also makes one pound boxes of MixedFroot, and although the recipe is different, each box still contains an integral number of ounces of the same four fruits. Finally, DryFroot produces giant boxes of TootyFroot-ies by mixing four pounds of Froot-Mix with three pounds of Mixed-Froot. In each of the three products, the weights of the four fruits are dif-ferent. The respective boxes list the ingredients in descending order by weight as follows: FrootMix: dates, cranberries, bananas, apples; Mixed-Froot: apples, bananas, cranberries, dates; TootyFrooties: bananas, cran-berries, dates, apples. How many ounces of each fruit are there in a box of each product?

—An Enigma by Keith Austin in New Scientist

5 Solve the following cryptic equa-tion: TBP = √(PUZZLE + 1) All the usual rules apply. There are no leading zeros. Each letter repre-sents a different digit, and the same

letter always represents the same digit.

—Adapted from Allan Gottlieb’s Puzzle Corner in Technology Review

Bonus A dozen POWs share a cell. What is important to them is that they escape as soon as possible. They have found a way to remove the bars on the window of their cell so they can climb through and, once on the ground, plan to head due north, climb over the east-west fence near the window, and disappear. But they can only hope to escape unseen if they do it in the dark, one man at a time, and with each man having from the time he starts, a clear two min-utes with all four guards at least 200 meters away from a point directly below their window. The guards each walk an east-west route, two inside the fence and two outside. Robinson starts 300 me-ters west of the prisoners’ window, marches to a point exactly outside their window, and turns around and marches back. Smith starts outside their window, marches 400 meters to the east, turns around, and marches back. Outside the fence, Phillips starts 500 meters west of their window, marches 550 meters to his turn around point, 50 meters east of their window. Quinn starts at Phillips’ turn around point, marches 500 me-ters east and turns around. The guards march back and forth on their beats at a uniform speed of 100 meters per minute. (Assume no time is spent in turning.) Because it is winter, the POWs reckon that it is dark enough for their purposes at 6 p.m. until 6 a.m. The guards of one shift begin their routes at the above indicated loca-tions at precisely 6 p.m. How many prisoners can success-fully escape? At what time(s) should they start their attempts? Express your answer in minutes.

—Brain Puzzler’s Delight by E.R. Emmet

Computer Bonus A k-digit au-tomorphic, or circular, number is a number whose square ends in the same k digits as the number itself.

45Summer 2015 The BenT of Tau BeTa pi

That is, N is a k-digit automorphic number if N2 ≡ N (mod 10k). For example, 76 is a 2-digit automorphic number because 762 = 5776. K-digit automorphic numbers are known to come in pairs. What is the leftmost (that is, most significant) digit of each of the two 1,001-digit, base-10, automorphic numbers?

—J.C. Rasbold, OH Α ’83

Send your answers to any or all of the Summer Brain Ticklers to Curt Gomulinski, Tau Beta Pi, P. O. Box 2697, Knoxville, TN 37901-2697 or email to BrainTicklers@tbp.org plain text only. The cutoff date for entries to the Summer column is the appearance of the Fall Bent in early October. It is not necessary to include the method of solution. The Computer Bonus is not graded. The judges welcome any interest-ing problems that might be suitable for the column. Curt forwards your entries to the judges, who are H. G. McIlvried III, PA Γ ’53, F. J. Ty-deman, CA D ’73, D. A. Dechman, TX Α ’57, and the columnist for this issue, J. C. Rasbold, OH Α ’83.

TELL US THE TALE...WIN A T-SHIRT

Send uS your clever caption(s) for this photo from The Bent archives, and if it is judged one of the best, you will win a TBP t-shirt.

The picture above, published in the Spring 1974 issue, was included in an article about Northrop University (at the time, Northrop Institute of Technology) where the California Pi Chapter was located. As part of an under-graduate senior research project investigating the effects of industrial noise on brain wave activity, the student above records the brain waves of a fellow student.

Email entries to pat@tbp.org, or mail them to HQ by August 10, 2015.

The photo for the Spring Contest, to the right, showed

guests staying to dance following the Michigan Regional Banquet held on April 5, 1974. It was attended by mem-bers, faculty, and guests of Michigan Alpha, Michigan Gamma, Michigan Zeta, and the Flint Alumnus Chapter.

From a field of 21 entries, this caption, submitted by David W. Kortebein, IL Α ’85, was #1 with the judges: “Yes dear, I do think that my new sport coat will still be in style 40 years from now.” Second place goes to Michael B. Dubey, NY E ’47, for his caption: “Really? I didn’t know that engineers could dance!” Congratulations to both winners who will receive a TBP t-shirt.

An entry submitted by Richard ”Lance” Null, AZ B ’90, is deserving of honorable mention: “Maintain 5 psi pressure on female. Right foot, left foot. Rotate right 90 degrees…”

Thanks to all of the participants for your witty com-mentaries. Keep those entries com-ing!!

GATES CAMBRIDGE SCHOLARS

Three Tau Bates have been named as Gates Cambridge Scholars among the 94 students chosen from around the world chosen for 2015. The schol-arships are full-cost awards to pur-sue a full-time postgraduate degree at the University of Cambridge.

Allison M. Kindig, IA B ’15, is an industrial en-gineering senior at the University of Iowa. She will pursue a M.Phil. in engineering for sustainable development at Cambridge. She is a 2014 Scholar,

2015 Fellow, and vice president of Iowa Beta. Kindig plans to work on helping communities achieve energy and food security.

Shruti Sharma, MA B ’15, who is majoring in materials science and engineering at MIT, will pursue a Ph.D. in engineering at Cambridge. She conducts research on the use of three dimensional printing for pros-

thetic prototyping and production. Sharma is the elected president of the MIT student body and founder of the Girls Lead-ership and Men-torship (GLAM) program.

Chiedozie Ibekwe, MS B ’11, plans to pursue a M.Phil. in public

policy and "will utilize my manu-facturing and supply chain man-agement expertise to advise African policymakers on crafting and ex-ecuting effective industrial policies

to boost manufacturing and diver-sify African economies." Ibekwe has been employed at GE, currently as a lead buyer at GE Oil & Gas, since graduating with a degree in electri-cal engineering.